Question: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 2} = \dfrac{-7x + 18}{x - 2}$
Explanation: Multiply both sides by $x - 2$ $ \dfrac{x^2}{x - 2} (x - 2) = \dfrac{-7x + 18}{x - 2} (x - 2)$ $ x^2 = -7x + 18$ Subtract $-7x + 18$ from both sides: $ x^2 - (-7x + 18) = -7x + 18 - (-7x + 18)$ $ x^2 + 7x - 18 = 0$ Factor the expression: $ (x + 9)(x - 2) = 0$ Therefore $x = -9$ or $x = 2$ At $x = 2$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 2$, it is an extraneous solution.